AP EAMCET · PHYSICS · Communication System
In a communication system operating at wavelength \(800 \mathrm{~nm}\), only one per cent of source frequency is available as signal bandwidth. The number of channels accommodated for transmitting TV signals of bandwidth \(6 \mathrm{MHz}\) is
- A \(\frac{1}{25} \times 10^7\)
- B \(\frac{1}{21} \times 10^{\prime}\)
- C \(\frac{1}{16} \times 10^7\)
- D \(\frac{1}{12} \times 10^7\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{16} \times 10^7\)
Step-by-step Solution
Detailed explanation
Given that, wavelength, \(\lambda=800 \mathrm{~nm}\) speed of light, \(c=3 \times 10^8 \mathrm{~m} / \mathrm{s}\) Then, frequency, \(f=c / \lambda\) Substituting the values, we get Source frequency, \(f=\frac{3 \times 10^8}{800 \times 10^{-9}} \mathrm{~Hz}\)…
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